2017 AMC 12B Problem 18

Question

The diameter

\overline{AB}

of a circle of radius

2

is extended to a point

D

outside the circle so that

BD = 3.

Point

E

is chosen so that

ED = 5

and line

ED

is perpendicular to line

AD.

Segment

\overline{AE}

intersects the circle at a point

C

between

A

and

E.

What is the area of

\triangle ABC?

\dfrac{120}{37}

\dfrac{140}{39}

\dfrac{145}{39}

\dfrac{140}{37}

\dfrac{120}{31}

\angle ACB

is inscribed in a semicircle, so it is a right angle; hence

\triangle ABC \sim \triangle AED

Their areas are in ratio

AB^2 : AE^2;

compute

AE^2 = AD^2 + ED^2

Since

\angle ACB

is inscribed in a semicircle, it is a right angle, so

\triangle ABC \sim \triangle AED

(both right-angled and sharing angle

A

). Their areas are in ratio

AB^2 : AE^2.

Here

AB = 4,

so

AB^2 = 16,

and

AD = AB + BD = 7,

so

AE^2 = AD^2 + ED^2

= 49 + 25

= 74.

The area of

\triangle AED

is

\tfrac12 \cdot 7 \cdot 5 = \tfrac{35}{2}.

Thus

[\triangle ABC] = \frac{16}{74} \cdot \frac{35}{2} = \frac{140}{37}.

Thus, the correct answer is D.

Problem 18 in other years